MCMC diagnostics and finite-run limitations — Examples

Each row of $P$ is nonnegative and sums to one. The uniform law is invariant because each column also sums to one, so $\pi P=\pi$ by [F1] and [F2]. By [F5], the transition rule is time homogeneous. Repeated use of the Markov property gives, for times $n_1<\cdots <n_r$ and states $x_1,\dots ,x_r$, the path probability $\pi (x_1){\prod }_{j=1}^{r-1}{P}^{n_{j+1}-n_j}(x_j,x_{j+1})$: the first marginal is $\pi$ because $\pi P=\pi$, and each later gap uses the corresponding matrix iterate. This expression depends only on the gaps, so a common time shift leaves every finite-dimensional law unchanged, which is strict stationarity by [F3].

Under $\pi$, $\mathbb{E}[Y_n]=0$ and $\mathrm{Var}(Y_n)=1$ by [F6]. Given $Y_n=1$, the next value is $1$ with probability $.9$ and $-1$ with probability $.1$; given $Y_n=-1$, the signs reverse. Thus $\mathbb{E}[Y_{n+1}\mid Y_n]=.8Y_n$. Iteration gives $\mathbb{E}[Y_{n+k}\mid Y_n]={.8}^k{Y}_n$. Since $\mathbb{E}{Y}_n=0$ and $\mathrm{Var}(Y_n)=1$, [F6] gives $\rho (k)=\mathbb{E}(Y_n{Y}_{n+k})={.8}^k$.

The sum is geometric, so $\tau =1+2{\sum }_{k\ge 1}{.8}^k=1+2(.8\mathrm{/}.2)=9$. By [F4], the ESS for $n$ unthinned transitions is $n\mathrm{/}9$.

Keeping every tenth state gives retained lag-$j$ correlation ${.8}^{10j}$. Therefore ${\tau }_{10}=1+2{\sum }_{j\ge 1}{.8}^{10j}=(1+{.8}^{10})\mathrm{/}(1-{.8}^{10})\approx 1.240$, and there are $n\mathrm{/}10$ retained states. Its ESS is approximately $(n\mathrm{/}10)\mathrm{/}1.240=.0806n$, less than $n\mathrm{/}9=.1111n$. The displayed lag-one correlation is lower, but the total information under a fixed transition budget is also lower.

The masses of $\pi$ sum to $1\mathrm{/}2+16(1\mathrm{/}32)=1$, so $\pi$ is a probability measure by [F3]. The maps $K$ and $\mathrm{\Pi }$ have probability rows by construction, so their convex mixture $P$ is also a transition matrix by [F1]. The kernel $K$ fixes the narrow state and permutes sixteen equally weighted broad states, hence $\pi K=\pi$. Every row of $\mathrm{\Pi }$ is $\pi$, hence $\pi \mathrm{\Pi }=\pi$; linearity gives $\pi P=.99\pi K+.01\pi \mathrm{\Pi }=\pi$.

By [F5], these are time-homogeneous chains with transition rule $P$. Each has 31 transitions. The independent transition choices all select $K$ with probability ${.99}^{4\cdot 31}$. On this event each chain completes two cycles through $B$ and never visits $n$, despite $\pi (\{n\})=1\mathrm{/}2$.

Each original chain has length 32 and traverses the same 16-state cycle twice. Splitting gives eight identical length-16 chains. For either coordinate, the rank-normalized split sequences are identical and nonconstant, so $B=0$ and $W>0$. The pooled medians are $2.5$ for $v$ and $0$ for $x$; the folded deviations each have at least two distinct values, so their split sequences are also identical and nonconstant.

With split length $N=16$ and $B=0$, the split formula gives ${\widehat{V}}^{+}=(15\mathrm{/}16)W$ for the rank-normalized and folded draws of both coordinates. Each statistic is $\sqrt{15\mathrm{/}16}$, so [F4] returns $\widehat{R}=\sqrt{15\mathrm{/}16}<1$. These values do not reveal the common missed state.

Put $h(x):=x^{2}f(x)=x^2\mathrm{/}(\pi (1+x^2))$, which is nonnegative and measurable by [F11]. For every finite $R>1$ and $x\in [1,R]$, $x^2\mathrm{/}(1+x^2)\ge 1\mathrm{/}2$, so $(2\pi {)}^{-1}{\mathbf{1}}_{[1,R]}(x)\le h(x)$. By [F9] and [F10], ${\int }_{\mathbb{R}}hd\lambda \ge {\int }_{\mathbb{R}}(2\pi {)}^{-1}{\mathbf{1}}_{[1,R]}d\lambda =\frac{R-1}{2\pi }\mathrm{.}$ These lower bounds are unbounded as $R\to \mathrm{\infty }$, so $\int hd\lambda =+\mathrm{\infty }$. Applying [F8] to the nonnegative measurable function $x\mapsto x^2$ gives ${\int }_{\mathbb{R}}{x}^{2}d{\mu }_C={\int }_{\mathbb{R}}{x}^{2}fd\lambda ={\int }_{\mathbb{R}}hd\lambda =+\mathrm{\infty }\mathrm{.}$ Thus the Cauchy target is not square-integrable, and its population variance is not defined under [F1].

Splitting the arrays gives four length-four chains, two with values $(-1,-1,1,1)$ and two with $(1,1,3,3)$. Each has sample variance $4\mathrm{/}3$; their means are $0,0,2,2$, with grand mean $1$. Hence $W=4\mathrm{/}3$ and $B=(4\mathrm{/}3)(1+1+1+1)=16\mathrm{/}3$.

In the pooled sample, $-1,1,3$ have multiplicities $4,8,4$ and average ranks $2.5,8.5,14.5$. Their Blom probabilities are $17\mathrm{/}130,1\mathrm{/}2,113\mathrm{/}130$. By [F7], the corresponding scores are $-a,0,a$ for $a={\mathrm{\Phi }}^{-1}(113\mathrm{/}130)>0$. The four split-chain means are $-a\mathrm{/}2,-a\mathrm{/}2,a\mathrm{/}2,a\mathrm{/}2$, each variance is $a^2\mathrm{/}3$, and the scale factors cancel in [F2]; the rank-normalized split statistic is again $\sqrt{7\mathrm{/}4}$.

Therefore ${\widehat{V}}^{+}=(3\mathrm{/}4)(4\mathrm{/}3)+(1\mathrm{/}4)(16\mathrm{/}3)=7\mathrm{/}3$, and the raw split statistic is $\sqrt{(7\mathrm{/}3)\mathrm{/}(4\mathrm{/}3)}=\sqrt{7\mathrm{/}4}$.

The pooled median is $1$, so the raw folds have values $0$ and $2$, each occurring eight times. Their average ranks are $4.5$ and $12.5$, with Blom probabilities $33\mathrm{/}130$ and $97\mathrm{/}130$. By [F7], their scores are $-b$ and $b$, where $b={\mathrm{\Phi }}^{-1}(97\mathrm{/}130)>0$. Thus every folded score is $b$ times its raw fold minus $b$. A positive scaling and common translation multiply $W$ and $B$ by $b^2$ and leave ${\widehat{V}}^{+}\mathrm{/}W$ unchanged. The raw folded split chains are $(2,2,0,0)$ and $(0,0,2,2)$; each has mean $1$ and sample variance $4\mathrm{/}3$, so $B=0$ and the rank-normalized folded statistic is $\sqrt{3\mathrm{/}4}$.

The maximum in [F3] is $\sqrt{7\mathrm{/}4}$. The calculation detects this finite location shift, while step 1.1 shows why variance-based population reasoning is unavailable for the Cauchy target.